Compute the self inductance

[leonne]

Homework Statement

calc the self inductance of a hairpin...to get a definite answer assume the wire has a tiny radius of e

Homework Equations

L=$$\phi$$/I

The Attempt at a Solution

So i found the b field $$\mu$$I/2$$\pi$$s
than for the flux, I got $$\mu$$I/2$$\pi$$$$\int$$1/s Lds
In the book, they got the same thing, but they had it 2$$\mu$$I/2$$\pi$$$$\int$$1/s Lds
Where did they get the 2 from? is it becasue we find the flux for both top wire and bottom wire?

 

[mark726]

Hello, thank you for your question! The reason for the 2 in the book's solution is because the hairpin is essentially a looped wire, with current flowing in both directions (clockwise and counterclockwise) along the two sides of the hairpin. This means that the magnetic field created by one side will cancel out the magnetic field created by the other side, resulting in a net magnetic field of 0. However, when calculating the self-inductance, we need to consider the flux created by both sides of the wire, so we multiply by 2. I hope this helps clarify the solution for you. Let me know if you have any further questions.